Model:Only the two currents enclosed by the closed path contribute to the line integral. Visualize:Please refer to Figure EX33.22. Solve:Ampere’s law gives the line integral of the magnetic field around the closed path: 50through1.3810T mB d sIμ−⋅==×GGú()()()70233410T m/A8.0 AIIIμπ−=+=×+⇒()5371.3810T m8.0 A410T m/AIπ−−×+=×⇒I3=3.0 A, out of the page. Assess:The right-hand rule was used above to assign a positive sign to I2. Since I3is also positive, it is in the same direction as I2.

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33.23.Model:The magnetic field is that of a current flowing into the plane of the paper. The current-carrying wire is very long. Visualize:Please refer to Figure EX33.23. Solve: Divide the line integral into three parts: fisemicircleright lineleft lineB d sB d sB d sB d s⋅=⋅+⋅+⋅∫∫∫∫GGGGGGGGThe magnetic field of the current-carrying wire is tangent to clockwise circles around the wire. BGis everywhere perpendicular to the left line and to the right line, thus the first and third parts of the line integral are zero. Along the semicircle, BGis tangent to the path andhas the same magnitude B=μ0I/2πdat every point. Thus 7f600i(410T m/A)(2.0 A)00()1.2610T m222IIBdsBLddμμπππ−−×⋅=++====×∫GGwhere L=πdis the length of the semicircle, which is half the circumference of a circle of radius d.

33.24.Model:Assume that the solenoid is an ideal solenoid. Solve:We can use Equation 33.16 to find the current that will generate a 3.0 mT field inside the solenoid: 0solenoidNIBlμ=⇒solenoid0BlINμ=Using l=0.15 m and 0.15 m0.0010 m150,N==()()()()()373.010T0.15 m2.4 A410T m/A150Iπ−−×==Assess:This is a reasonable current to pass through a good conducting wire of diameter 1 mm.

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33.25.Model:Assume that the solenoid is an ideal solenoid. Solve:We can use Equation 33.16 to find the current that will generate a magnetic field of 1.5 T inside the solenoid: 0solenoidNIBlμ=⇒solenoid0BlINμ=Using l=1.8 m and 1.8 m0.0020900,N==()()()71.5 T1.8 m2.4 kA410T m/A900Iπ−==Assess:Large currents can be passed through conducting wires. The current density through the superconducting wire is ()()2822400 A0.001 m7.610A/mπ=×. This value is reasonable for a superconducting wire.

33.26.Model:A magnetic field exerts a magnetic force on a moving charge. Visualize:Please refer to Figure EX33.26. Solve:(a)The force on the charge is ()()()()()()()()197on q71913ˆˆˆ1.6010C1.010m/scos45sin450.50T 1.010m/sˆˆˆˆˆ1.6010C0.50 T5.710N2FqvBikiiikij−−−=×=××° +°××=××+×= +×GGG(b)Because the cross product ˆˆii×in the equation for the force is zero, on q0 N.F=GG

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